If the error in the measurement of radius of a sphere is `2%`, then the error in the determination of volume of the sphere will be

Options

A. 0.08
B. 0.02
C. 0.04
D. 0.06

Answer

D

Explanation

`V=4/3piR^(3),"In"V=In(4/3pi)+InR^(3)`
Differentiating `(dV)/V=3(dR)/R`,Error in the determination of the volume `=3xx2%=6%`